You have found the following ages (in years) of 5 tigers. Those tigers were randomly selected from the 28 tigers at your local zoo: $ 4,\enspace 23,\enspace 5,\enspace 26,\enspace 10$ Based on your sample, what is the average age of the tigers? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 28 tigers, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{4 + 23 + 5 + 26 + 10}{{5}} = {13.6\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {92.16} + {88.36} + {73.96} + {153.76} + {12.96}} {{5 - 1}} $ {s^2} = \dfrac{{421.2}}{{4}} = {105.3\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{105.3\text{ years}^2}} = {10.3\text{ years}} $ We can estimate that the average tiger at the zoo is 13.6 years old. There is also a standard deviation of 10.3 years.